Gas pressure and Brownian motion
Use collisions to explain pressure, and discover how the motion of a visible speck gives clues about invisible particles.
Core: gas collisions and Brownian observations/explanations. Supplement / Extended: collision forces per unit area and precise distinctions between molecules and larger specks. This is Cambridge IGCSE, not Egyptian Baccalaureate / Thanaweya Amma or Year 3.
A gas pushes because its particles reach the wall
Gas particles move continuously in random directions. They collide with surfaces and rebound. Each collision contributes a tiny push; vast numbers of wall collisions produce the pressure we measure. Collisions between particles can change their motion, but pressure on a wall comes directly from collisions with that wall.
Worked example
Air in a bicycle tyre pushes against the inside surface because moving air particles hit it repeatedly. Air outside also pushes on the outside. The tyre’s behaviour depends on the difference between inside and outside pressures as well as the rubber.
Think first: Do particles need to fall downwards to produce gas pressure?
No. They move in all directions, so a stationary gas can press on the sides and top as well as the base.
Build an explanation as a chain
- Name the particles: gas molecules or a larger suspended speck?
- State the observation and the conditions held fixed.
- Describe the relevant collisions: with a wall for gas pressure; with a speck for Brownian motion.
- Link frequency, strength or imbalance to the outcome; do not say particles simply become bigger.
- For a pressure calculation, distinguish one-face force from a net force using two pressures.
Worked Brownian explanation: four connected points
The speck is struck by moving particles of the surrounding fluid [1]. Their collision effects do not exactly balance [1]. The imbalance changes unpredictably, giving irregular motion [1]. This supports continuous random motion in the surrounding fluid even though its molecules are not resolved by the microscope [1]. “The speck is alive” or “specks collide with each other” does not describe this mechanism.
Worked calculation: pressure difference across a lid
Inside pressure is 110000 Pa, outside pressure is 100000 Pa, and area is 0.020 m². Inside force = 2200 N outward; outside force = 2000 N inward. Net pressure force = (110000 − 100000) × 0.020 = 200 N outward. Assume a flat lid and uniform pressures; weight and other forces are excluded. Using inside pressure alone would answer a different question.
Gas pressure laboratory
Predict, then explain the collisions
Core activity. Keep the amount of gas fixed, with no leaks or change of state. Each scenario states which other condition stays fixed. No home heating or compression experiment is needed.
Heat a sealed rigid vessel
Unequal-collision laboratory
Which way is the unbalanced push?
Core idea: collisions need not balance. This simplified model counts equal-strength collision effects during one short interval. Every left-side hit contributes one rightward “push unit”; right-side hits contribute leftward units, and similarly for top and bottom. These units are not newtons, and real collisions vary in strength. The result describes the combined collision effect, not a complete prediction of the speck’s displacement.
Snapshot 1 of 8. The large circle is the suspended speck. Arrows show incoming collision effects, not light rays.
Changing snapshots clears your prediction. Supplement connection: light, fast-moving molecules can affect a larger speck because vast numbers collide and their momentum transfers fluctuate.
Practise, then explain
Four multiple-choice questions, three numerical responses and five written tasks. The numerical questions revisit Core pressure skills from section 1.8.1. Detailed collision-force reasoning and precise speck/molecule distinctions are Supplement / Extended. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.
Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.
Which event directly produces the gas pressure on a container wall?
Optional hint
The question asks about a force exerted on the surface, not interactions elsewhere.
Source clarifications and safe observations
The original PDF is unchanged. A force is a push, but pressure is force per unit area. The relation p = F/A is Core in section 1.8.1; applying microscopic collision-force reasoning is Supplement in 2.1.2.7. Our numerical recall questions keep that distinction clear.
The source calls invisible collisions the “only explanation” for a moving speck. More carefully, controlled Brownian observations support the particle model; currents, vibration and stage movement must be considered. A shared drift does not rule out Brownian motion superimposed on it. Do not infer a quantitative particle-mass rule from one speck or one short track.
Expansion comparisons here explicitly keep temperature and amount of gas fixed. Forces calculated from one pressure are not automatically net forces across a lid. Figure 21.1 inspired the schematic collision and sampled-path activities; their examples are not real experimental measurements. Use a safe teacher-prepared cell or video, never smoke generation or heated sealed containers at home.
Source alignment: supplied Physics-0625 Volume A, PDF pp.149–154 / printed pp.141–146, Figure 21.1; official Cambridge 2026–2028 syllabus, section 2.1.2. Original learning text, activities and questions. Existing source-download access is unchanged.